Whirling of shafts — a disc fixed at mid-span has CG eccentricity e = 2 mm relative to the shaft axis. The lateral natural frequency is 10 rad/s. If the shaft rotates at 300 r.p.m., the steady whirl radius will be approximately:
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A2 mm
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B2.22 mm
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C2.50 mm
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D3.0 mm
Answer
Correct Answer: 2.22 mm
Explanation
Introduction / Context: An unbalanced rotor (disc with eccentric center of gravity) mounted on a flexible shaft will whirl at a radius that depends on running speed relative to the lateral natural frequency. This is the classical Jeffcott (Laval) rotor model.
Given Data / Assumptions:
- Eccentricity e = 2 mm = 0.002 m.
- Undamped lateral natural frequency ωn = 10 rad/s.
- Running speed 300 r.p.m. ⇒ ω = 2π×300/60 = 10π ≈ 31.416 rad/s.
- Neglect damping for the amplitude estimate.
Concept / Approach: For an undamped Jeffcott rotor, steady whirl amplitude X equals X = e·ω²/|ωn² − ω²|. Since ω > ωn here, use X = e·ω²/(ω² − ωn²).
Step-by-Step Solution:
1) Compute ω = 31.416 rad/s; ω² ≈ 985.96.2) Compute ω² − ωn² ≈ 985.96 − 100 = 885.96.3) Evaluate X = 0.002 × 985.96 / 885.96 ≈ 0.002225 m = 2.225 mm.4) Round suitably → 2.22 mm.Verification / Alternative check: Limiting cases: as ω→ωn, amplitude grows large (resonance); as ω ≫ ωn, X→e (approaches the geometric eccentricity), which aligns with the computed value being slightly above 2 mm.
Why Other Options Are Wrong:2 mm — would be the asymptote as ω ≫ ωn, but ω is only ~3.14× ωn.2.50 mm and 3.0 mm — overestimate the dynamic amplification at this speed ratio.
Common Pitfalls: Using (ωn² − ω²) in the denominator for ω > ωn; forgetting to convert r.p.m. to rad/s.
Final Answer: 2.22 mm.